# Using the identity theorem: can there be an analytic function $f$ with $f\left(\frac{1}{n^2}\right) = \frac{1}{n}$

As Conway states it the theorem is as follows:

Let $G$ be an open connected set and let $f:G\rightarrow \mathbb{C}$ is analytic on $G$. Then the TFAE:

1. $f\equiv0$
2. $\{z\in G: f(z)=0 \}$ has a limit point in $G$.

I get confused when I have to use this to solve problems. My understanding on this is sort of like "if $f$ goes to zero along a certain sequence" then the function must be identically zero. Is this thinking correct? Even if I am I want some good explanation on this. Also, if any of you could give a good reference to this that would help too.

For example what can we say about an analytic function $f:\mathbb C\rightarrow \mathbb C$, such that $f\left(\frac{1}{n^2}\right) = \frac{1}{n}$. Can such a function exist? Can we even use the identity theorem to answer this question? Thanks so much for your time.

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"if f goes to zero along a certain sequence" isnt right, what the theorem says is that if the function $f=0$ on a set which accumlates in its domain, then the function will be identically equal to zero on the domain. – TTY Jan 5 '13 at 6:06
@user54755: I thought that I should share another method with you. By way of contradiction, assume that $f$ exists. We have $\dfrac{1}{n^{2}} \to 0$ and $\dfrac{1}{n} \to 0$, so by continuity, $f(0) = 0$. Next, we attempt to find $f'(0)$ the old-fashioned way: $\displaystyle \lim_{n \to \infty} \dfrac{f(1/n^{2}) - f(0)}{1/n^{2}} = \lim_{n \to \infty} \dfrac{(1/n) - 0}{1/n^{2}} = \lim_{n \to \infty} n$. This says that $f'(0)$ does not exist, which is a contradiction. – Haskell Curry Jan 5 '13 at 7:44
@user54755: Strictly speaking, your use of $f(z) = \sqrt{z}$ is not entirely correct. I know that you want to identify $f(z)$ with $\sqrt{z}$ on the set $\left\{ \dfrac{1}{n} \right\}_{n \in \mathbb{N}} \cup \{ 0 \}$, but notice that $\sqrt{z}$ is not analytic on any open subset of $\mathbb{C}$ that contains $0$. This is because $0$ is a branch point for $\sqrt{z}$. I think this is perhaps why Michael hesitated a little on your comment. But good attempt anyway! :) – Haskell Curry Jan 5 '13 at 7:57
Thanks @HaskellCurry. – Jack Dawkins Jan 5 '13 at 17:51

What you said about $f$ going to zero along a sequence isn't quite correct. Consider $a_n = \frac{1}{n}$ and $f(z) = z$. We have $f\left(\frac{1}{n}\right) \to 0$ as $n \to \infty$, but $f$ is not identically zero. However, if $f : G \to \mathbb{C}$ is holomorphic (analytic), and there is a sequence ${a_n}$ in $G$ with $a_n \to a \in G$, and $f(a_n) = 0$ for all $n$, then $f(a) = 0$ (by continuity), and hence the set $\{z \in G : f(z) = 0\}$ contains the limit point $a$. Therefore $f$ is identically zero.
As for your second question, suppose $f$ exists and consider the holomorphic function $z \mapsto f(z^2) - z$. What are its values at $a_n = \frac{1}{n}$? What does that tell us about $f$?
Oh okay. So the new function (call it $g$) has the property that $g(a_n)=0$ for all $n$. Since f is holomorphic so is g. And $a_n \rightarrow 0$. So by continuity it follows that $g(0)=0$ and therefore by idenity theorem $g$ must be identically zero. But this shows that $f(z^2)=z$ for all $z$. Therefore $f(z)=\sqrt z$, but this function is not analytic at zero and thus no such function can exist. – Jack Dawkins Jan 5 '13 at 6:20
Pretty much. A more elementary way to see that $f$ doesn't exist is to note that $1 = f(1^2) = f(1)$ and $-1 = f((-1)^2) = f(1)$, which of course can't happen. – Michael Albanese Jan 5 '13 at 6:26